---
title: "Vertical curve (-)3% and (-)35%, IP 2477 m"
subject: "Surveying II"
chapter: 6
key: "semester_4/surveying_ii:vertical-curve-neg3-neg35-ip-2477"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-neg3-neg35-ip-2477
site: BCT Engineering
---

# Vertical curve (-)3% and (-)35%, IP 2477 m

Surveying II · Chapter 6: Curves

- Asked in: 2079 Chaitra, 6 marks
- Video: [FE Exam Transportation - Vertical Curve Problem 1](https://www.youtube.com/watch?v=5IhcpnHoA1o) — Genie Prep
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-neg3-neg35-ip-2477
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6

## Question

A grade of (-)3% meets another grade of (-)35%. The elevation and chainage of IP are 2477.00 m and 4500 m respectively. Field conditions require that the vertical curve should pass through a point of elevation 2474.50 m at chainage 4500.00 m. Compute a suitable equal tangent vertical curve including full station elevations. Take peg interval 30 m.

## Short answer

Using the mid-ordinate e = 2.5 m from IP to curve, length L = 800×2.5/32 = 62.5 m. Equal-tangent summit curve: BVC at 4468.75 m (2477.938 m), EVC at 4531.25 m (2466.062 m). Highest point BVC, lowest EVC.

## Answer

The second grade is taken as written, $-35\%$. (If it was meant to be $-3.5\%$, the same method gives $L = 800\times 2.5/0.5 = 4000$ m.)

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve.

#### Step 1: Length of the curve
The curve is a summit curve, so the curve lies below the IP:

$$
e = 2474.5 - 2477 = -2.5\ \text{m}\ \Rightarrow\ |e| = 2.5\ \text{m}
$$

For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence

$$
L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 2.5}{|-35-(-3)|} = \frac{2000}{32} = 62.5\ \text{m}
$$

#### Step 2: BVC, EVC and their levels
The curve is a summit (crest) curve with equal tangent lengths $L/2 = 31.25$ m.

$$
\begin{aligned}
\text{Ch. of BVC} &= 4500 - 31.25 = 4468.75\ \text{m}\\
\text{Ch. of EVC} &= 4500 + 31.25 = 4531.25\ \text{m}\\
\text{RL of BVC} &= 2477 - \frac{(-3)}{100}\times 31.25 = 2477.938\ \text{m}\\
\text{RL of EVC} &= 2477 + \frac{(-35)}{100}\times 31.25 = 2466.062\ \text{m}
\end{aligned}
$$

#### Step 3: RL of curve points
Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$
y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}
$$

with $g_1 = -3\%$, $g_2 = -35\%$, $L = 62.5$ m, so the offset is $y = \frac{(-32)x^2}{200\times 62.5}$ and pegs are taken at multiples of 30 m.

| Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) |
|---|---|---|---|---|
| 4468.75 | 0 | 2477.938 | 0.000 | 2477.938 |
| 4470 | 1.25 | 2477.900 | -0.004 | 2477.896 |
| 4500 | 31.25 | 2477.000 | -2.500 | 2474.500 |
| 4530 | 61.25 | 2476.100 | -9.604 | 2466.496 |
| 4531.25 | 62.5 | 2476.062 | -10.000 | 2466.062 |

Check: the last curve RL (2466.062 m) equals the RL of EVC (2466.062 m), and the curve at the IP chainage lies 2.5 m below the IP, at RL 2474.5 m.

#### Step 4: Highest and lowest points
Both grades are falling, so the grade never becomes zero within the curve and there is no turning point; the highest level is at the BVC (RL 2477.938 m) and the lowest level is at the EVC (RL 2466.062 m).

**Answer:** $L = 62.5$ m, BVC at 4468.75 m (RL 2477.938 m), EVC at 4531.25 m (RL 2466.062 m); curve levels as tabulated.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The curve passes 2.5 m below the IP. This means it is
   A. undefined
   B. a sag curve
   C. a straight
   D. a summit (crest) curve
   Answer: D. A curve lying below the IP is a summit curve; the grade change is from -3% to -35%, a downward bend. (Key point: Summit vs sag)

2. With e = 2.5 m and change of grade 32%, L =
   A. 62.5 m
   B. 31.25 m
   C. 125 m
   D. 250 m
   Answer: A. L = 800 x 2.5/32 = 62.5 m. (Key point: Length)

3. The BVC chainage with IP 4500 m is
   A. 4531.25 m
   B. 4468.75 m
   C. 4500 m
   D. 4437.5 m
   Answer: B. 4500 - L/2 = 4500 - 31.25. (Key point: BVC chainage)

4. Why is there no turning point?
   A. e is negative
   B. Summit curves never have one
   C. Both grades are falling, so the grade does not become zero
   D. The curve is too short to measure
   Answer: C. Both grades are negative, so the highest level is at BVC and the lowest at EVC. (Key point: Turning point)

## Close variants set in other exams

- [Vertical curve 5% and 3%, 1474.50 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-5-3-ip-1475) — Surveying II, chapter 6, asked 7×
- [Vertical curve +3.5% and -0.5%, 1266 m at 780](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35down-05-1266-780) — Surveying II, chapter 6, asked 8×
- [Vertical curve -2.5% and +3.5%, 1268.50 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-25down-35up-12685) — Surveying II, chapter 6, asked 9×
- [Vertical curve +0.5% and -3.5%, 1266 m at 780](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-05up-35down-1266-at) — Surveying II, chapter 6, asked 9×
- [Vertical curve +3.5% and -2.75%, 30 m pegs](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35-down-275-through) — Surveying II, chapter 6, asked 7×

## Related in Surveying II

- [Vertical curve -3.5% and +0.5%, IP 1300 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg35-up-05-1300) — Surveying II, chapter 6, asked 5×
- [Vertical curve -3.5% and +5%, 1268 m at 780](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35down-5up-1268-at) — Surveying II, chapter 6, asked 3×
- [Vertical curve -3.5% and -0.5%, 1268 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg35-neg05-1268-lowest) — Surveying II, chapter 6
- [Vertical curve +1.25% and +4.75%, 1518.30 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-125-475-151760-2030) — Surveying II, chapter 6, asked 7×
- [Vertical curve, -3.5% and +0.5%, 3268 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-parabolic-why-3267-1780) — Surveying II, chapter 6, asked 3×
- [Vertical curve +1.25% and +4.75%, 1267.70 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-125-475-126770) — Surveying II, chapter 6, asked 3×

## Related in other subjects

- [Vertical curve +4% and -3%, through given point](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70m-vc-4-3) — Transportation Engineering I, chapter 3
- [Vertical curve 35 m headlight and momentum grade](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-80c-vc-headlight-35m) — Transportation Engineering I, chapter 3
- [Vertical curve chainage and elevations, R 8000 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-69bh-vc-chainage) — Transportation Engineering I, chapter 3
- [Vertical curve 5% and 4% at 50+050](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-78p-vc-5-4) — Transportation Engineering I, chapter 3